Codeforces Round #105 (Div. 2) C. Terse princess

C. Terse princess

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

«Next please», — the princess called and cast an estimating glance at the next groom.

The princess intends to choose the most worthy groom, this is, the richest one. Whenever she sees a groom who is more rich than each of the previous ones, she says a measured «Oh...». Whenever the groom is richer than all previous ones added together, she exclaims «Wow!» (no «Oh...» in this case). At the sight of the first groom the princess stays calm and says nothing.

The fortune of each groom is described with an integer between 1 and 50000. You know that during the day the princess sawn grooms, said «Oh...» exactly a times and exclaimed «Wow!» exactly b times. Your task is to output a sequence ofn integers t₁, t₂, ..., t_n, wheret_i describes the fortune ofi-th groom. If several sequences are possible, output any of them. If no sequence exists that would satisfy all the requirements, output a single number-1.

Input

The only line of input data contains three integer numbers n, a and b (1 ≤ n ≤ 100, 0 ≤ a, b ≤ 15, n > a + b), separated with single spaces.

Output

Output any sequence of integers t₁, t₂, ..., t_n, wheret_i (1 ≤ t_i ≤ 50000) is the fortune ofi-th groom, that satisfies the given constraints. If no sequence exists that would satisfy all the requirements, output a single number-1.

Sample test(s)

Input

10 2 3

Output

5 1 3 6 16 35 46 4 200 99

Input

5 0 0

Output

10 10 6 6 5

Note

Let's have a closer look at the answer for the first sample test.

• The princess said «Oh...» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99.
• The princess exclaimed «Wow!» (highlighted in bold): 5 1 3 6 16 35 46 4 200 99.

题目链接：http://codeforces.com/problemset/problem/148/C

题目大意：n个数的数组，如果当前数字比它前面所有的数字都大，则是情况Oh，情况Oh的数有a个。如果这个数比它前面所有数字的和都大，则是情况Wow，情况Wow的数有b个。构造这样的数组。如果数组不存在，输出-1。

解题思路：二进制思维处理Wow的情况，1 2 4 8 16......每个数都比前面的数字和大，先构造Wow的情况，再构造Oh的情况，最后构造一般情况。注意：1.b=0且普通情况只有一个时不能构造；2.第一个数是普通情况；3.Wow,Oh条件都满足时时Wow情况。

代码如下：

#include <cstdio>
#include <cstring>
int t[105];
int main(void)
{
	int n,a,b;
	scanf("%d%d%d",&n,&a,&b);
	if(!b&&a&&n-a==1)
	{
		printf("-1\n");
		return 0;
	}
	printf("1");
	t[0]=1;
	for(int i=1;i<n;i++)
	{
		if(!b&&i==1)
		{
			t[1]=1;
			printf(" 1");
			a++;
			continue;
		}
		if(i<=b)
			t[i]=1<<i;
		else if(i<=a+b)
			t[i]=t[i-1]+1;
		else
			t[i]=t[i-1];
		printf(" %d",t[i] );
	}
	printf("\n");
}